Question Video: Solving Logarithmic Equations Involving Laws of Logarithms and Quadratic Equations Mathematics

Solve logβ‚‚ (log₃ (π‘₯Β² βˆ’ 8π‘₯)) = 1, where π‘₯ ∈ ℝ.

02:31

Video Transcript

Solve log to the base two of log to the base three of π‘₯ squared minus eight π‘₯ equals one where π‘₯ is in the set of real numbers.

So the first thing we can do with this problem is make sure that we’ve got a log to the same base on each side of the equation. And we can do that by applying one of our log rules. And that is that log to the base π‘Ž of π‘Ž equals one. So, therefore, we can say that log to the base two of log to the base three of π‘₯ squared minus eight π‘₯ equals log to base two of two because, as we said, one is just the same as log to the base two of two.

So now the reason we’ve done this is because what we’ve got is log to the same base on the left- and right-hand sides. So, what we can do is equate our arguments. So, we can say that log to the base three of π‘₯ squared minus eight π‘₯ is equal to two. So now, to solve for π‘₯, what we can do first is actually rearrange from logarithmic to exponent form. So if we have log to the base π‘Ž of 𝑏 equals π‘₯, then we say that π‘Ž to the power of π‘₯ is gonna be equal to 𝑏.

So therefore, if we identify our π‘Ž, 𝑏, and π‘₯, well then we can rewrite our equation as three squared equals π‘₯ squared minus eight π‘₯, which is gonna give us nine equals π‘₯ squared minus eight π‘₯. So then, what we’re gonna do is subtract nine from each side of the equation. So, we get zero equals π‘₯ squared minus eight π‘₯ minus nine. So now what we need to do is solve this quadratic.

Well, we can solve the quadratic by factoring. And if we do, we get zero equals π‘₯ minus nine multiplied by π‘₯ plus one. So therefore, we can say that π‘₯ is gonna be equal to nine or negative one. So great, and these both are going to be in our solution set. Well, you might think, β€œWell, no, it can’t be because we know that the argument must be positive and not equal to one.” So therefore, we cannot have the negative value. However, that is not the case in this particular problem because if we take a look at the argument, the argument is not π‘₯. The argument is π‘₯ squared minus eight π‘₯, well, the argument of one of our logarithms.

And what we’re gonna do to show that is actually substitute in π‘₯ equals negative one into this argument. Well, if we do that, we get negative one all squared minus eight multiplied by negative one, which is gonna give us one plus eight because negative one squared is just one. And if you subtract a negative, it’s the same as adding a positive. Well, this gives us the value of nine, which is both positive and not equal to one. So therefore, it satisfies the conditions that we have for our argument. So therefore, we can say that the solution set for our equation is nine and negative one.

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